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Aldrich Ames is a convicted traitor who leaked American secrets to a foreign power. Yet Ames took routine lie detector tests and each time passed them. How can this be done? Recognizing control questions, employing unusual breathing patterns, biting one's tongue at the right time, pressing one's toes hard to the floor, and counting backward by 7 are countermeasures that are difficult to detect but can change the results of a polygraph examination (Source: Lies' Liesi't Lies\%' The Psychology of Deceit, by C. V. Ford, professor of psychiatry, University of Alabama). In fact, it is reported in Professor Ford's book that after only 20 minutes of instruction by "Buzz" Fay (a prison inmate), \(85 \%\) of those trained were able to pass the polygraph examination even when guilty of a crime. Suppose that a random sample of nine students (in a psychology laboratory) are told a "secret" and then given instructions on how to pass the polygraph examination without revealing their knowledge of the secret. What is the probability that (a) all the students are able to pass the polygraph examination? (b) more than half the students are able to pass the polygraph examination? (c) no more than four of the students are able to pass the polygraph examination? (d) all the students fail the polygraph examination?

Step by step solution

01

Define the Context and Variables

We are dealing with a probability problem involving a binomial distribution. The probability of each student passing the polygraph test, given the training, is 85% or \( p = 0.85 \). The number of students, \( n \), is 9.

02

Calculate the Probability of All Passing

To find the probability that all 9 students pass the exam, we use the formula for binomial probability where \( k = 9 \): \[ P(X = 9) = \binom{9}{9} (0.85)^9 (0.15)^0 \]. The calculation is \( P(X = 9) = (0.85)^9 \approx 0.247 \).

03

Calculate the Probability of More than Half Passing

More than half means 5 to 9 students pass. We calculate the sum \( \sum_{k=5}^{9} P(X = k) \). Using the formula for each individual probability, you compute: \[ P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) \]. The result is approximately 0.995.

04

Calculate the Probability of No More Than Four Passing

No more than four passing means \( P(X \leq 4) \). Compute \( P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \) using the binomial formula for each. The sum is approximately 0.0049.

05

Calculate Probability That All Fail

The probability that all students fail means none pass: \( X = 0 \). Use \( P(X = 0) = \binom{9}{0} (0.15)^9 \approx 3.84 \times 10^{-6} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Probability Theory

At the core of this problem is probability theory, an intriguing branch of mathematics that attempts to quantify uncertainty. Probability theory helps us understand and calculate the chances of different outcomes happening. In this context, we are dealt with a problem involving nine students where we need to calculate the probability of different scenarios:

• All students passing the polygraph examination.
• More than half the students passing.
• No more than four students passing.
• All students failing.

When we talk about probability, we're dealing with the likelihood of a single event occurring. These probabilities can range from 0, meaning zero chance, to 1, meaning absolute certainty. In this exercise, we have a specific probability of a student passing the polygraph test, set at 85% or 0.85. By understanding these basic probabilities, we can dive deeper into statistical methods to solve more complex scenarios.

Applying Statistical Methods

Statistical methods allow us to interpret and analyze probability in a structured manner. In this exercise, we apply a statistical model called the binomial distribution. It is perfectly fitted for experiments or scenarios where there are only two possible outcomes—such as passing or failing a test.
For example, to find the likelihood that all nine students pass, we use the binomial probability formula: \[ P(X = 9) = \binom{n}{k} (p)^k (1-p)^{n-k} \]where:

• \( n \) is the number of trials (students in this case),
• \( k \) is the number of successes (students passing),
• \( p \) is the probability of passing a single trial.

We further expand this method to find the probability of more than five passing and no more than four passing by summing the individual probabilities for each possibility. These methods are foundations in statistics, providing a powerful toolset for analyzing outcomes in a predictable way.

Polygraph Examination and Countermeasures

The subject of a polygraph examination revolves around detecting lies through physiological responses. However, there are notorious accounts of individuals bypassing these tests, one of which involves Aldrich Ames, a traitor who managed to pass routine lie detector tests. This leads to a detailed exploration of countermeasures in polygraph examinations.
Countermeasures refer to tactics that individuals might use to manipulate results. These can include deliberate actions like controlling breathing patterns, biting one's tongue, or counting backward. These actions aim to interfere with the physiological responses that the polygraph machine measures.
This exercise reflects on the unsettling efficacy of such techniques. According to Professor Ford, simple instructions delivered by an individual named "Buzz" Fay were enough for a significant number of participants to pass the polygraph test knowingly. The exercise introduces an amazing 85% pass rate among those employing these techniques, sparking genuine discussions on the reliability of polygraph tests. It's an interesting intersection of psychology and statistical methods, reminding us that numbers alone don't always tell the whole story.